1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Step2247 [10]
3 years ago
11

On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around

the curve is 7,500 Newtons. What speed limit should be posted for the curve so that cars can negotiate it safely
Physics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

You might be interested in
A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to
NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) =  (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0
3 years ago
Transverse waves are generally stronger than longitudinal waves. True or False ?
kondaur [170]
False
If all other factors, such as medium, are kept the same, longitudinal waves tend to be stronger.
8 0
3 years ago
1. Tonya had a hard time deciding between the Big Burger and the Crispy Chicken sandwiches, her two favorites.
Radda [10]
Bro the md that she lost was the md boneless burger
8 0
3 years ago
If you saw a waxing gibbous moon what phase would you expect one week later?
Leona [35]
Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
6 0
3 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 22 km/h and the coef
Aleks [24]
First, let's put 22 km/h in m/s:

22 \frac{km}{h} \times  \frac{1000m}{1km}  \times  \frac{1h}{3600s}=6.11 \frac{m}{s}

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

F_{rad}=m \frac{v^2}{R}

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

F_{fric}= mg \mu_{s}

Notice we don't use the kinetic coefficient even though the bike is moving.  This is because when the tires meet the road they are momentarily stationary with the road surface.  Otherwise the bike is skidding.

Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:

m\frac{v^2}{R} = mg \mu_{s} \\ \\ \frac{v^2}{R} = g \mu_{s} \\ \\R= \frac{v^2}{g \mu_{s}} \\ \\ R= \frac{6.11}{9.8 \times 0.37}=1.685m

3 0
3 years ago
Other questions:
  • Agility refers to a person's level of flexibility.
    5·1 answer
  • A tow can pull a car out of a ditch in 7.5 seconds. How many work is done if the truck has power of 5500 watts?
    6·1 answer
  • Two waves equal amplitudes meet other result in a wave with zero amplitudes. Which phenomenon
    6·1 answer
  • Which statement best defines work.
    6·1 answer
  • 23. Challenge: A gas is heated so that it expands from a volume of 1.0L to a volume of 1.5 l. If the
    13·1 answer
  • How does an iron nail become a magnet when it is placed in a strong magnetic field
    15·2 answers
  • A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co
    15·1 answer
  • Rafael lives near a road at the bottom of the hill. His parents are concerned that soil will wash off the hill and rocks will fa
    6·1 answer
  • Visible light waves do not diffract as well as radio waves because
    9·1 answer
  • Can you please help me​
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!